Verified answer

Объяснение:

[tex]sin70^0-cos40^0=sin70^0-sin(90^0-40)=sin70^0-sin50^0=\\=2*sin\frac{70^0-50^0}{2}*cos\frac{70^0+50^0}{2}=2*sin10^0*cos60^0=2*sin10^0*\frac{1}{2}=sin10^0.[/tex]

Ответ:

sin70

0

−cos40

0

=sin70

0

−sin(90

0

−40)=sin70

0

−sin50

0

=

=2∗sin

2

70

0

−50

0

∗cos

2

70

0

+50

0

=2∗sin10

0

∗cos60

0

=2∗sin10

0

∗

2

1

=sin10

0

.